Third term in the expansion of `(3x - (y^3)/(6) ) ` is ………..

Answer

To find the third term in the expansion of \( (3x - \frac{y^3}{6})^4 \), we will use the Binomial Theorem.

### Step-by-step Solution:

1. **Identify the Binomial Expansion**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
Here, we have \( a = 3x \), \( b = -\frac{y^3}{6} \), and \( n = 4 \).

2. **Determine the Third Term**:
The \( r \)-th term in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For the third term, we need \( r = 2 \) (since \( T_{3} \) corresponds to \( r = 2 \)).

3. **Calculate the Binomial Coefficient**:
We need to calculate \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

4. **Substitute Values into the Formula**:
Now we substitute \( n = 4 \), \( r = 2 \), \( a = 3x \), and \( b = -\frac{y^3}{6} \):
\[ T_3 = \binom{4}{2} (3x)^{4-2} \left(-\frac{y^3}{6}\right)^2 \]
This simplifies to:
\[ T_3 = 6 (3x)^{2} \left(-\frac{y^3}{6}\right)^{2} \]

5. **Calculate Each Component**:
- Calculate \( (3x)^{2} = 9x^{2} \)
- Calculate \( \left(-\frac{y^3}{6}\right)^{2} = \frac{y^6}{36} \)

6. **Combine the Results**:
Now substitute back into the term:
\[ T_3 = 6 \cdot 9x^{2} \cdot \frac{y^{6}}{36} \]
Simplifying this gives:
\[ T_3 = \frac{54x^{2}y^{6}}{36} = \frac{3}{2}x^{2}y^{6} \]

### Final Answer:
The third term in the expansion of \( (3x - \frac{y^3}{6})^4 \) is:
\[ \frac{3}{2} x^{2} y^{6} \]